The Angle at the Centre

Pick an arc (or chord) AB on a circle. It is "seen" from two places: from the centre O, and from any point P out on the circle. These two viewing angles are not equal — there is a beautifully simple relationship between them. The angle at the centre is exactly double the angle at the circumference:

\angle AOB = 2 \times \angle APB

This is the master circle theorem — most of the other circle-angle facts fall out of it as special cases.

For an arc (or chord) AB of a circle with centre O:

the arc subtends an angle at the centre \angle AOB, and an angle at the circumference \angle APB at any point P on the circle (on the major arc);
the angle at the centre is exactly double the angle at the circumference — \angle AOB = 2\,\angle APB;
this is the master circle theorem: angles in the same segment, the angle in a semicircle, and more all come straight from it.

Why it works

It rests on one fact you already have: every radius is the same length, so the triangles inside are isosceles. Their equal base angles, added through the exterior-angle rule, give the doubling. Step through it.

In symbols: in isosceles \triangle OAP the base angles are a, and the exterior angle at O is 2a; in \triangle OBP they are b, exterior angle 2b. So \angle AOB = 2a + 2b = 2(a + b) = 2\,\angle APB.